In the coordinate plane, the curve $xy = 1$ intersects a circle at four points, three of which are $\left( 2, \frac{1}{2} \right),$ $\left( -5, -\frac{1}{5} \right),$ and $\left( \frac{1}{3}, 3 \right).$  Find the fourth point of intersection.
Answer: Let the equation of the circle be $(x - a)^2 + (y - b)^2 = r^2.$  From $xy = 1,$ $y = \frac{1}{x}.$  Substituting, we get
\[(x - a)^2 + \left( \frac{1}{x} - b \right)^2 = r^2.\]Then
\[x^2 - 2ax + a^2 + \frac{1}{x^2} - \frac{2b}{x} + b^2 = r^2,\]so
\[x^4 - 2ax^3 + (a^2 + b^2 - r^2) x^2 - 2bx + 1 = 0.\]By Vieta's formulas, the product of the roots is 1.  Three of the roots are 2, $-5,$ and $\frac{1}{3},$ so the fourth root is $-\frac{3}{10}.$  Therefore, the fourth point is $\boxed{\left( -\frac{3}{10}, -\frac{10}{3} \right)}.$